Sherlock and Squares | HackerRank

This is an easy hackerrank challenge which will help you to become good at competitive programming. There are various competitive programming websites like CodeChef, HackerEarth, Codeforces where you can practice coding.

Watson likes to challenge Sherlock’s math ability. He will provide a starting and ending value describing a range of integers. Sherlock must determine the number of square integers within that range, inclusive of the endpoints.

Note: A square integer is an integer which is the square of an integer, e.g. 1, 4, 9, 16, 25.

For example, the range is a = 24 and b = 49, inclusive. There are three square integers in the range: 25, 36 and 49.

Read full problem : Sherlock and Squares.

Let the integer variables first_num store the first number of the range and last_num store the last number of the range.

To find number of square integers in the range [first_num, last_num], first we calculate square root of the first_num and an integer variable count stores the number of square integers in the range.

int sr = sqrt(first_num);
int count = 0;

For eg. in the range [35, 70], variable sr is equal to 5 since √35 = 5.916. If square of sr is greater than or equal to the first_num and less than or equal to the last_num then the number (sr * sr) is in the range and is a square integer.

while ((sr * sr) <= last_num)
{
    if ((sr * sr) >= first_num)
        count++;
    sr++;
}

#include <iostream>
#include <cmath>

int num_of_squares(int first_num, int last_num) {
    int sr = sqrt(first_num);
    int count = 0;
    while ((sr * sr) <= last_num)
    {
        if ((sr * sr) >= first_num)
            count++;
        sr++;
    }
    return count;
}

int main()
{
    int test_cases;
    std::cin >> test_cases;

    while (test_cases--)
    {
        int first_num, last_num;
        std::cin >> first_num >> last_num;

        std::cout << num_of_squares(first_num, last_num) << "\n";
    }
}

View this code on Github.

If you have another solution then please share it in the comments below.